# Number System Practice Gym by Anubhav Sehgal - Part 2

• Q1) Find the sum of all possible values of x such that 2/(sqrt(x)) + 1/(sqrt(y)) = 1/(sqrt(2)) where x and y are positive integers

u = sqrt(x) and v = sqrt(y)
2v + u = uv/rt(2)
uv - 2rt2 v - rt2 u = 0
v(u - 2rt2) - rt2(u - 2rt2) = 4
(v - rt2)(u - 2rt2) = 4 = rt2 * 2rt2 = 2rt2 * rt2
u,v = 4rt2,2rt2 ; 3rt2,3rt2
x,y = 32,8 ; 18,18
Sum = 50

Q2) How many three digit numbers are 34 times of sum of their sum of digits?

100a + 10b + c = 34a + 34b + 34c
66a - 24b - 33c = 0
66a = 24b + 33c
a = 1 ; b = 0 ; c = 2
a = 2 ; b = 0 ; c = 4
a = 3 ; b = 0 ; c = 6
a = 4 ; b = 0 ; c = 8
Four such numbers.

Q3) Find the number of integers, N such that (N^2 + 2N - 8 )/(N^2 + N - 12) is an integer.

N^2 + 2N - 8 = (N - 2)(N + 4)
N^2 + N - 12 = (N - 3)(N + 4)
=> (N - 2)/(N - 3) is an integer
=> (N - 3 + 1)/(N - 3) is an integer
=> 1 + 1/(N - 3) is an integer => N = 4. Only one possible value.

Q4) Find the 1000th term of the sequence : 1,3,4,7,8,9,10,11,13,14,... in which there is no number which contain digit 2,5 or 6.

I have two approaches for this one :
The Shareef approach :smile:
The Tareef approach :stuck_out_tongue_winking_eye:

The Shareef approach :
_ : 6 numbers ; _ _ : 6 * 7 = 42 numbers ; _ _ _ : 6 * 7 * 7 = 294 numbers
1 _ _ _ : 7 * 7 * 7 = 343 added = 685 numbers till now. Remaining 1000 - 685 = 315
3 0 _ _ : 49 added
3 1 _ _ : 49 added
3 3 _ _ : 49 added
3 4 _ _ , 3 7 _ _ , 3 8 _ _ = 49*3 = 147
Total 294 more. Remaining 21
3 9 0 _ : 7
3 9 1 _ : 7
3 9 3 _ : 7
3939

The Tareef approach :
0 -> 0
1 -> 1
2 -> 3
3 -> 4
4 -> 7
5 -> 8
6 -> 9
1000th term : 1000 in base 7 = 2626 3939

Q5) How many positive divisors of 100^10 end in exactly two zeroes?

100^10 = 2^20 * 5^20
2^18 * 5^18
19 + 19 - 1 = 37

Q6) 960 = 880
c + 880 + c + 800 + c = 1680 + 3c > 1000 Not to be considered
OR
for x c = 1 to 13
x = -21 to 8
Option c)

Q7) How many pairs of non negative integers exist,such that the difference between their product and sum is 72?

ab - a - b = 72
ab - a - b + 1 = 73
a(b - 1) - 1(b - 1) = 73
(a - 1)(b - 1) = 73 = 1 * 73 = -1 * -73
a,b = 2,74 ; 74,2 ; 0,-72 ; -72,0
Non negative integer pairs = 1 unordered or 2 ordered
Total integral solutions = 2 unordered or 4 ordered

Q8) The equation 3^(x - 1) + 5^(x - 1) = 34 has :
a) No solution
b) 1 solution
c) 2 solutions
d) More than 2 solutions

34 = 25 + 9
x = 3 is the only solution
Both are exponentially increasing.
3 + 5 = 8< 34 ; 9 + 25 = 34 ; 27 + 125 > 34. No other values. Option b)

Q9) For what smallest positive integer N, will the number 3! + 5! + 7! + .. + (2N + 1)! be a perfect square ?

3!(1 + 4 * 5 + 4 * 5 * 6 * 7 + ....) = 3! * odd . Power of 2 is 1. So no possible N.

Q10) What is the unit digit of [10^3000/(10^100 + 3)] where [.] is GIF.

Remainder = 10^3000/(10^100 + 3)
Put x = 10^100 => x^30/x + 3. Put x = -3 => (-3)^30 = 3^30
Dividend = Quotient*Divisor + Remainder
Check for unit digit on both sides,
0 = x * 3 + 9
x = 7

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